Trigonometric equation (急) 20分

2009-05-30 5:32 am
solve the following Trigonometric equation︰

sin²x+3cos²x-2cosx-1=0
更新1:

2. ︱ cosx︱= (開方3)/2

回答 (3)

2009-05-30 5:35 am
✔ 最佳答案
sinx + 3cosx - 2cosx - 1 = 0

(sinx + cosx) + 2cosx - 2cosx - 1 = 0

1 + 2cosx - 2cosx - 1 = 0 (Recall the identity sinx + cosx = 1)

2cosx - 2cosx = 0

cosx(cosx - 1) = 0

cosx = 0 or cosx = 1

x = 90* or 270* or x = 0* or 360* (For 0* <= x <= 360*)

Combined solution: x = 0*, 90*, 270* or 30*


2009-05-30 09:14:30 補充:
│cosx│ = sqrt3 / 2

cosx = sqrt3 / 2 or -sqrt3 / 2

x = 30* or 330* or x = 150* or 210*

Solution: x = 30*, 150*, 210* or 330*

2009-05-30 12:15:44 補充:
下面的朋友漏了一些solution
參考: Physics king
2009-05-30 6:45 pm
1. sin&sup2;x+3cos&sup2;x-2cosx-1=0
sin&sup2;x+cos&sup2;x+2cos&sup2;x-2cosx-1=0
1+2cos&sup2;x-2cosx-1=0
2cos&sup2;x=2cosx
cosx=0
x=90(degree)

2.︱ cosx︱= (開方3)/2
either cosx =(開方3)/2 or -cosx=(開方3)/2
x=30(degree) or x=330(degree)
if cosx is negative, then
x=150(degree) or x=210(degree)
2009-05-30 5:51 am
︱ cosx︱= (開方3)/2
cosx=(開方3)/2 或 cosx=(-開方3)/2
x=30度,150度,210度或330度


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